2. Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
Dot Product of Vectors
3. Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
4. Example A: Let u = <5, -6>, v = <4, -5>
a. u v
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
5. Example A: Let u = <5, -6>, v = <4, -5>
a. u v = <5, -6> <4, -5>
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
6. Example A: Let u = <5, -6>, v = <4, -5>
a. u v = <5, -6> <4, -5>
= 5*4 + (-6)*(-5)
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
7. Example A: Let u = <5, -6>, v = <4, -5>
a. u v = <5, -6> <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
8. Example A: Let u = <5, -6>, v = <4, -5>
a. u v = <5, -6> <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
b. 3u v = 3<5, -6> <4, -5>
9. Example A: Let u = <5, -6>, v = <4, -5>
a. u v = <5, -6> <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
b. 3u v = 3<5, -6> <4, -5>
= <15, -18> <4, -5>
10. Example A: Let u = <5, -6>, v = <4, -5>
a. u v = <5, -6> <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
b. 3u v = 3<5, -6> <4, -5>
= <15, -18> <4, -5>
= 15*4 + (-18)*(-5)
= 60 + 90 = 150
11. Example A: Let u = <5, -6>, v = <4, -5>
a. u v = <5, -6> <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
b. 3u v = 3<5, -6> <4, -5>
= <15, -18> <4, -5>
= 15*4 + (-18)*(-5)
= 60 + 90 = 150 = 3(u v)
12. Example A: Let u = <5, -6>, v = <4, -5>
a. u v = <5, -6> <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
b. 3u v = 3<5, -6> <4, -5>
= <15, -18> <4, -5>
= 15*4 + (-18)*(-5)
= 60 + 90 = 150 = 3(u v)
Facts: 1. u•v = v•u
13. Example A: Let u = <5, -6>, v = <4, -5>
a. u v = <5, -6> <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
b. 3u v = 3<5, -6> <4, -5>
= <15, -18> <4, -5>
= 15*4 + (-18)*(-5)
= 60 + 90 = 150 = 3(u v)
Facts: 1. u•v = v•u
2. For any scalar λ, λu • v = λ(u • v)
14. Example A: Let u = <5, -6>, v = <4, -5>
a. u v = <5, -6> <4, -5>
= 5*4 + (-6)*(-5)
= 20 + 30 = 50
Given u=<a, b>, v=<c, d>, the dot product of u
and v is u v = <a, b> <c, d> = ac + bd
The dot product of two vector is a number.
Dot Product of Vectors
b. 3u v = 3<5, -6> <4, -5>
= <15, -18> <4, -5>
= 15*4 + (-18)*(-5)
= 60 + 90 = 150 = 3(u v)
Facts: 1. u•v = v•u
2. For any scalar λ, λu • v = λ(u • v)
3. u•(v + w) = u•v + u•w
15. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Unitized Vectors
16. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
Unitized Vectors
17. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
Unitized Vectors
18. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
Unitized Vectors
19. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
b: Let λ = , then the length of <3, 4> is1
5
1
5
Unitized Vectors
20. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
b: Let λ = , then the length of <3, 4> is
|< , >| = * |<3, 4>|
1
5
1
5
3
5
4
5
1
5
Unitized Vectors
21. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
b: Let λ = , then the length of <3, 4> is
|< , >| = * |<3, 4>| = * 5 = 1
1
5
1
5
3
5
4
5
1
5
1
5
Unitized Vectors
22. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
b: Let λ = , then the length of <3, 4> is
|< , >| = * |<3, 4>| = * 5 = 1
1
5
1
5
3
5
4
5
1
5
1
5
Vector of length 1
are called unit vectors.
Unitized Vectors
23. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
b: Let λ = , then the length of <3, 4> is
|< , >| = * |<3, 4>| = * 5 = 1
1
5
1
5
3
5
4
5
1
5
1
5
Vector of length 1
are called unit vectors.
<3/5, 4/5>
r = 1
Unit vectors
Unitized Vectors
24. Fact: Given v a vector, λ a scalar, then |λv| = |λ|*|v|.
Example B: Let v = <3, 4> so that | <3, 4> | = 5,
a: Let λ = 10, then the length of 10<3, 4> is
|<30, 40>| = 10 * 5 = 50
b: Let λ = , then the length of <3, 4> is
|< , >| = * |<3, 4>| = * 5 = 1
1
5
1
5
3
5
4
5
1
5
1
5
Vector of length 1
are called unit vectors.
In the above example,
3
5
4
5
< , > is a unit vector.
<3/5, 4/5>
r = 1
Unit vectors
Unitized Vectors
26. Unitized Vectors
In the above exampe b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.3
5
4
5,
27. Unitized Vectors
In the above exampe b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.
We say that we unitized v.
3
5
4
5,
28. Unitized Vectors
In the above exampe b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.
We say that we unitized v.
3
5
4
5,
From this example, we see that in general
the unitized v is 1
|v|
v
29. Unitized Vectors
In the above exampe b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.
We say that we unitized v.
3
5
4
5,
From this example, we see that in general
the unitized v is since | v | = 1.1
|v|
1
|v| |v| =
1
|v|
v
30. Unitized Vectors
In the above exampe b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.
We say that we unitized v.
3
5
4
5,
Geometrically, unitizing v
corresponds to shrinking
or stretching v to be a
unit vector.
From this example, we see that in general
the unitized v is since | v | = 1.1
|v|
1
|v| |v| =
1
|v|
v
31. Unitized Vectors
In the above exampe b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.
We say that we unitized v.
3
5
4
5,
From this example, we see that in general
the unitized v is since | v | = 1.1
|v|
1
|v| |v| =
1
|v|
v
Geometrically, unitizing v
corresponds to shrinking
or stretching v to be a
unit vector.
v
r=1
unitized v
32. Unitized Vectors
In the above exampe b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.
We say that we unitized v.
3
5
4
5,
Geometrically, unitizing v
corresponds to shrinking
or stretching v to be a
unit vector.
v
1
|v|
v
r=1
unitized v
unitized v
,
From this example, we see that in general
the unitized v is since | v | = 1.1
|v|
1
|v| |v| =
1
|v|
v
33. Unitized Vectors
In the above exampe b. we rescale the vector
v =<3, 4> and obtain < > which is a unit vector.
We say that we unitized v.
3
5
4
5,
Geometrically, unitizing v
corresponds to shrinking
or stretching v to be a
unit vector.
v
1
|v|
v
r=1
unitized v
unitized v
,
Unitized v is the only unit
vector that has the same
direction as v.
From this example, we see that in general
the unitized v is since | v | = 1.1
|v|
1
|v| |v| =
1
|v|
v
35. Dot product and Angles
The following theorem put all the above concepts
together.
36. Dot product and Angles
Dot Product Theorem:
The following theorem put all the above concepts
together.
37. Dot product and Angles
Dot Product Theorem:
a. Let u and v be two unit vectors and be
the angle between them, then cos() = u•v.
The following theorem put all the above concepts
together.
38. Dot product and Angles
Dot Product Theorem:
a. Let u and v be two unit vectors and be
the angle between them, then cos() = u•v.
b. Let u and v be any two vectors and be
the angle between them
cos() = (unitized u) (unitized v)
The following theorem put all the above concepts
together.
39. Dot product and Angles
Dot Product Theorem:
a. Let u and v be two unit vectors and be
the angle between them, then cos() = u•v.
b. Let u and v be any two vectors and be
the angle between them
cos() = (unitized u) (unitized v)
= v
|v|
u
|u|
Easy to memorize
The following theorem put all the above concepts
together.
40. Dot product and Angles
Dot Product Theorem:
a. Let u and v be two unit vectors and be
the angle between them, then cos() = u•v.
b. Let u and v be any two vectors and be
the angle between them
cos() = (unitized u) (unitized v)
= =v
|v|
u
|u|
u•v
|u|*|v|
Easy to memorize Easier way to compute
The following theorem put all the above concepts
together.
41. Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
42. uv=10,
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
43. uv=10, |u|=10,
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
44. uv=10, |u|=10, |v|=20,
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
45. uv=10, |u|=10, |v|=20, hence,
cos() = =
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
46. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
47. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So = cos-1(1/2)
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
48. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
49. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
50. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>,
51. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
52. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
(u + v)•(u – v) = 5 – 15 = -10
53. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
(u + v)•(u – v) = 5 – 15 = -10
|u + v| = 50,
54. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
(u + v)•(u – v) = 5 – 15 = -10
|u + v| = 50, |u – v| = 10
55. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
(u + v)•(u – v) = 5 – 15 = -10
|u + v| = 50, |u – v| = 10
So cos(A) = -10/500
56. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
(u + v)•(u – v) = 5 – 15 = -10
|u + v| = 50, |u – v| = 10
So cos(A) = -10/500 -0.447
57. uv=10, |u|=10, |v|=20, hence,
cos() = = = 10/200 = 1/2
So = cos-1(1/2) = 45o
Dot product and Angles
Example C:
a. Let u = <3, -1> and v = <2, -4>. Find the angle
between them.
u•v
|u|*|v|
10
10*20
b. Find the angle A between u + v and u – v.
u + v = <5, -5>, u – v = <1, 3>
(u + v)•(u – v) = 5 – 15 = -10
|u + v| = 50, |u – v| = 10
So cos(A) = -10/500 -0.447
A = cos-1(-0.447) 117o.
58. Dot product and Angles
Theorem: Given u, v two nonzero vectors such that
u•v=0, then u and v are perpendicular (orthorgonal).
59. Dot product and Angles
Theorem: Given u, v two nonzero vectors such that
u•v=0, then u and v are perpendicular (orthorgonal).
If is the angle between them and u•v = 0,
then cos() = = 0.
u•v
|u|*|v|
60. Dot product and Angles
Theorem: Given u, v two nonzero vectors such that
u•v=0, then u and v are perpendicular (orthorgonal).
If is the angle between them and u•v = 0,
then cos() = = 0.
Hence = cos-1(0) = 90o
u•v
|u|*|v|
61. Example D: <5, -3>•<3, 5> = 15 – 15 = 0,
hence they are orthogonal.
Dot product and Angles
Theorem: Given u, v two nonzero vectors such that
u•v=0, then u and v are perpendicular (orthorgonal).
If is the angle between them and u•v = 0,
then cos() = = 0.
Hence = cos-1(0) = 90o
u•v
|u|*|v|
63. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).
64. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).
65. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).projv (u)
66. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).projv (u)
Let < 1800 be the angle between u and v,
the signed length of the projv(u) is |u |*cos().
67. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).projv (u)
If is less than 90o then
cos() > 0, so |u|cos() is
also positive.
Let < 1800 be the angle between u and v,
the signed length of the projv(u) is |u |*cos().
68. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).projv (u)
If is less than 90o then
cos() > 0, so |u|cos() is
also positive. This means
the shadow is cast in the
direction of v.
Let < 1800 be the angle between u and v,
the signed length of the projv(u) is |u |*cos().
69. Projections
The projection of the
vector u onto the vector v
is the vertically cast
shadow-vector of u onto v.
It is denoted as projv (u).projv (u)
If is less than 90o then
cos() > 0, so |u|cos() is
also positive. This means
the shadow is cast in the
direction of v.
u
v
|u|cos() is positive means the
projv(u) is in the direction of v
< 90o
Let < 1800 be the angle between u and v,
the signed length of the projv(u) is |u |*cos().
70. Projections
If is more than 90o then
cos() < 0, so |u|cos() is
negative.
71. Projections
If is more than 90o then
cos() < 0, so |u|cos() is
negative. We get the
negative length of projv(u)
and projv(u) is cast in the
opposite direction of v.
72. Projections
If is more than 90o then
cos() < 0, so |u|cos() is
negative. We get the
negative length of projv(u)
and projv(u) is cast in the
opposite direction of v.
u
v
|u|cos() is negative so projv(u)
is in the opposite direction of v.
>90o
73. Projections
If is more than 90o then
cos() < 0, so |u|cos() is
negative. We get the
negative length of projv(u)
and projv(u) is cast in the
opposite direction of v.
u
v
|u|cos() is negative so projv(u)
is in the opposite direction of v.
>90o
74. Projections
If is more than 90o then
cos() < 0, so |u|cos() is
negative. We get the
negative length of projv(u)
and projv(u) is cast in the
opposite direction of v.
u
v
|u|cos() is negative so projv(u)
is in the opposite direction of v.
Example A: Given that
the angle between u and v
is 67o and |u|=23, |v|=28.
Find the length of the
shadow projv(u). Draw.
>90o
75. Projections
If is more than 90o then
cos() < 0, so |u|cos() is
negative. We get the
negative length of projv(u)
and projv(u) is cast in the
opposite direction of v.
u
v
|u|cos() is negative so projv(u)
is in the opposite direction of v.
Example A: Given that
the angle between u and v
is 67o and |u|=23, |v|=28.
Find the length of the
shadow projv(u). Draw.
|u|=23
67o
|v|=28
>90o
76. Projections
If is more than 90o then
cos() < 0, so |u|cos() is
negative. We get the
negative length of projv(u)
and projv(u) is cast in the
opposite direction of v.
u
v
|u|cos() is negative so projv(u)
is in the opposite direction of v.
Example A: Given that
the angle between u and v
is 67o and |u|=23, |v|=28.
Find the length of the
shadow projv(u). Draw.
|u|=23
67o
|v|=28
The length of projv(u) = |u|cos() = 23cos(67) 8.99
>90o
77. Projections
Example B: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
78. Projections
Example B: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
|u|=23
117o |v|=28
79. Projections
Example B: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
|u|=23
117o |v|=28
80. Projections
Example B: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
|u|=23
117o |v|=28
The length of proju(v) = |v|cos()
81. Projections
Example B: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
|u|=23
117o |v|=28
The length of proju(v) = |v|cos()
= 28cos(117) -12.7
82. Projections
Example B: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
|u|=23
117o |v|=28
The length of proju(v) = |v|cos()
= 28cos(117) -12.7
Note the negative answer indicates the shadow is
cast in the opposite direction of u.
83. Projections
Example B: Given that the
angle between u and v is
117o and |u|=23, |v|=28.
Find the length of the
shadow proju(v). Draw.
|u|=23
117o |v|=28
The length of proju(v) = |v|cos()
= 28cos(117) -12.7
Note the negative answer indicates the shadow is
cast in the opposite direction of u.
Your Turn: Given the angle
between u and v is 135o and
|u|=36, |v|=18. Draw and find
the length of the projv(u).
135o
|u|=36
v
Ans: Signed length of projv(u) = –36/2 –25.4
projv(u)
85. Theorem: Given two vectors u and v in the
standard position, the signed length of projv(u) is
u dotted with unitized v.
Projv(u)
u
v
Signed Length of Standard Projection
86. Theorem: Given two vectors u and v in the
standard position, the signed length of projv(u) is
u dotted with unitized v. That is, the signed length
of projv(u) is u•
v
|v|
Projv(u)
u
v
Signed Length of Standard Projection
87. Theorem: Given two vectors u and v in the
standard position, the signed length of projv(u) is
u dotted with unitized v. That is, the signed length
of projv(u) is u• or .
Projv(u)
v
|v|
u•v
|v|
u
v
Signed Length of Standard Projection
88. Theorem: Given two vectors u and v in the
standard position, the signed length of projv(u) is
u dotted with unitized v. That is, the signed length
of projv(u) is u• or .
v
|v|
u•v
|v|
Example C:
Let u=<3, -3> and
v=<-1, -4>. Find the length
of projv(u). Draw.
Projv(u)
u
v
Signed Length of Standard Projection
89. Theorem: Given two vectors u and v in the
standard position, the signed length of projv(u) is
u dotted with unitized v. That is, the signed length
of projv(u) is u• or .
Projv(u)
v
|v|
u•v
|v|
Example C:
Let u=<3, -3> and
v=<-1, -4>. Find the length
of projv(u). Draw.
u=<3, -3>
v=<-1, -4>
Signed Length of Standard Projection
90. Theorem: Given two vectors u and v in the
standard position, the signed length of projv(u) is
u dotted with unitized v. That is, the signed length
of projv(u) is u• or .
v
|v|
u•v
|v|
Example C:
Let u=<3, -3> and
v=<-1, -4>. Find the length
of projv(u). Draw.
uv = <3,-3><-1,-4> = 9
Projv(u)
u=<3, -3>
v=<-1, -4>
Signed Length of Standard Projection
91. Theorem: Given two vectors u and v in the
standard position, the signed length of projv(u) is
u dotted with unitized v. That is, the signed length
of projv(u) is u• or .
v
|v|
u•v
|v|
Example C:
Let u=<3, -3> and
v=<-1, -4>. Find the length
of projv(u). Draw.
uv = <3,-3><-1,-4> = 9
|v|= 17,Projv(u)
u=<3, -3>
v=<-1, -4>
Signed Length of Standard Projection
92. Theorem: Given two vectors u and v in the
standard position, the signed length of projv(u) is
u dotted with unitized v. That is, the signed length
of projv(u) is u• or .
Projv(u)
v
|v|
u•v
|v|
u
v
Example C:
Let u=<3, -3> and
v=<-1, -4>. Find the length
of projv(u). Draw.
uv = <3,-3><-1,-4> = 9
|v|= 17, hence the length of
projv(u) is =
Projv(u)
u=<3, -3>
v=<-1, -4>
u•v
|v|
9
17
Signed Length of Standard Projection
93. Theorem: Given two vectors u and v in the
standard position, the signed length of projv(u) is
u dotted with unitized v. That is, the signed length
of projv(u) is u• or .
Projv(u)
v
|v|
u•v
|v|
u
v
Example C:
Let u=<3, -3> and
v=<-1, -4>. Find the length
of projv(u). Draw.
uv = <3,-3><-1,-4> = 9
|v|= 17, hence the length of
projv(u) is =
Projv(u)
u=<3, -3>
v=<-1, -4>
u•v
|v|
9
17
2.18.
Signed Length of Standard Projection
96. Projections of Standard Vectors
For any vector v, v = (its's length)*(unitized v)
Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)
97. Projections of Standard Vectors
For any vector v, v = (its's length)*(unitized v)
Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)= vu•v
|v|2
98. Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)=
Projections of Standard Vectors
For any vector v, v = (its's length)*(unitized v)
Example D:
Let u=<3, -3> and
v=<-1, -4>. Find projv(u).
vu•v
|v|2
99. Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)=
Projections of Standard Vectors
u=<3, -3>v=<-1, -4>
For any vector v, v = (its's length)*(unitized v)
Example D:
Let u=<3, -3> and
v=<-1, -4>. Find projv(u).
vu•v
|v|2
100. Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)=
Projv(u)
Projections of Standard Vectors
Projv(u)
u=<3, -3>v=<-1, -4>
For any vector v, v = (its's length)*(unitized v)
Example D:
Let u=<3, -3> and
v=<-1, -4>. Find projv(u).
vu•v
|v|2
101. Projv(u)
Projections of Standard Vectors
Projv(u)
u=<3, -3>v=<-1, -4>
For any vector v, v = (its's length)*(unitized v)
Example D:
Let u=<3, -3> and
v=<-1, -4>. Find projv(u).
We've u•v = 9 and |v|2 =17.
Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)= vu•v
|v|2
102. Projv(u)
Projections of Standard Vectors
Projv(u)
u=<3, -3>v=<-1, -4>
For any vector v, v = (its's length)*(unitized v)
Example D:
Let u=<3, -3> and
v=<-1, -4>. Find projv(u).
We've u•v = 9 and |v|2 =17.
So proju(v) = vu•v
|v|2 *
Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)= vu•v
|v|2
103. Projv(u)
Projections of Standard Vectors
Projv(u)
u=<3, -3>v=<-1, -4>
For any vector v, v = (its's length)*(unitized v)
Example D:
Let u=<3, -3> and
v=<-1, -4>. Find projv(u).
We've u•v = 9 and |v|2 =17.
So proju(v) = vu•v
|v|2 *
9
17
* <-1, -4>=
Theorem: Given u and v in the standard position,
then the projection-vector in the standard form
projv(u)=(it's signed length)*(unitized v)= vu•v
|v|2
104. HW
Given u=<-1,2>, v=<3,-2>, and w=<4,2>
find the signed length and the vectors for
each of the following:
1. proju(v) 2. projv(u) 3. projv(w)
4. proju(u+v) 5. projw(v – u)
6. proju+v (u – w) 7. projv+w (v + u)